Equation of plane given 3 points

Asked Nov 21 '20 at 09:13

Active Nov 21 '20 at 09:13

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I am trying to find the equation of a plane that passes through the points $A = (0, 3 , 6)$, $B = (7, 1, -6)$ and $C = (-6, -9, -6)$.

I have done the cross product of $\overrightarrow{AB}$ and $\overrightarrow {AC}$ and obtained $120i + 156j - 96k$

Is that the final equation of the plane or am I missing a step?

I also want to find the distance from that plane to a point but I don't know how to do that, any pointers?

asked Nov 21 '20 at 09:13

SgZQoy9jhG

1

One vector does not specify a plane, however this vector is the vector normal to the plane and from this you can easily find the equation for the plane. E.g. look at step 3 in the answer to this post: https://math.stackexchange.com/questions/2686606/equation-of-a-plane-passing-through-3-points – Emiel Lanckriet Nov 21 '20 at 09:20
Do I have to divide my equation by the greatest common factor or can I substitute directly? – SgZQoy9jhG Nov 21 '20 at 09:26
That doesn't really matter, it will give you the same equation, bacause $lax + lby + lcz + lk = 0 \iff ax + by + cz + k = 0$. – Emiel Lanckriet Nov 21 '20 at 09:28
So the final equation would be $120x + 156y - 96z -108 = 0$ All three points should yield the same $k$, correct? – SgZQoy9jhG Nov 21 '20 at 09:31
Fill in the points and check. – Emiel Lanckriet Nov 21 '20 at 09:34
I am getting different answers for each, I must have done something wrong? – SgZQoy9jhG Nov 21 '20 at 09:35
Yes, you should get $0$ for all three points. – Emiel Lanckriet Nov 21 '20 at 09:35
1

The cross product appears to have different signs to what I am getting. I have equal signs for $i,k$-parts and opposite for the $j$-part. – String Nov 21 '20 at 10:32
3

Your cross product should either be (120, -156, 96) or (-120, 156, -96). Somewhere mistakes in signs. – Math Lover Nov 21 '20 at 11:47

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